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3 years ago

9l9-8xl=2x+3 check for extraneous solutions

Mathematics

2 answers:

iris [78.8K]3 years ago

6 0

9-(9-8x)=2x+3

simplify -72x+81=2x+3

subtract 2x from both sides -74x+81=3

subtract 81 from both sides -74x=-78

divide by -74 on both sides x=39/37

Kay [80]3 years ago

5 0

What level is this?
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The function a(t)=t^(1/2)−t^(−1/2) m/s^2 represents the acceleration of a particle moving along a horizontal axis. At time t=0,

garri49 [273]

Answer:

see below

Step-by-step explanation:

a(t)=t^(1/2)−t^(−1/2)

We integrate to find the velocity

v(t) = integral t^(1/2)−t^(−1/2) dt

= t ^ (1/2 +1) t ^ (-1/2 +1)

------------ - ----------------- + c where c is the constant of integration

3/2 1/2

v(t) = 2/3 t^ 3/2 - 2 t^ 1/2 +c

We find c by letting t=0 since we know the velocity is 4/3 when t=0

v(0) = 2/3 0^ 3/2 - 2 0^ 1/2 +c = 4/3

0+c =4/3

c = 4/3

v(t) = 2/3 t^ 3/2 - 2 t^ 1/2 +4/3

To find the position function we need to integrate the velocity

p(t) = integral 2/3 t^ 3/2 - 2 t^ 1/2 +4/3 dt

2/3 t ^ (3/2 +1) 2 t ^ (1/2 +1) 4/3t

------------ - ----------------- + ------------- + c

5/2 3/2 1

p(t) = 4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t +c

We find c by letting t=0 since we know the position is -4/15 when t=0

p(0) = 4/15 0^ 5/2 - 4/3 0 ^ 3/2 + 4/3*0 +c = -4/15

0 +c = -4/15

c = -4/15

p(t) = 4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t -4/15

8 0

3 years ago

Read 2 more answers

I think I know how to solve this, but could someone please check my work for me?

Fiesta28 [93]

Answer is :

cos⁻¹[(-1-4√7)/(√17, √8) = 173.33° ≈ 173°

5 0

3 years ago

Segment AB has endpoints at A(-7,2) and b(9,-6). Point P lies on AB such that AP:BP=3:1

bekas [8.4K]

The coordinates of the point P which divides the line segment AB made by the points A(-7,2) and B(9,-6) is (x,y) = (5,-4)

<h3>What is the coordinate of the point which divides a line segment in a specified ratio?</h3>

Suppose that there is a line segment $\overline{AB}$ such that a point P(x,y) lying on that line segment $\overline{AB}$ divides the line segment $\overline{AB}$ in m:n, then, the coordinates of the point P is given by:

$(x,y) = \left( \dfrac{mx_2 + nx_1}{m+n} , \dfrac{my_2 + ny_1}{m+n} \right)$

where we have:

the coordinate of A is $(x_1, y_1)$
and the coordinate of B is $(x_2, y_2)$

We're given that:

Coordinate of A is $(x_1, y_1)$ = (-7,2)
Coordinate of B is $(x_2, y_2)$ = (9.-6)
The point P lies on AB such that AP:BP=3:1 (so m = 3, and n = 1)

Let the coordinate of P be (x,y), then we get the values of x and y as:

$(x,y) = \left( \dfrac{mx_2 + nx_1}{m+n} , \dfrac{my_2 + ny_1}{m+n} \right)\\\\(x,y) = \left( \dfrac{3(9) + 1(-7)}{3+1} , \dfrac{3(-6) + 1(2)}{3+1} \right)\\\\(x,y) = \left( \dfrac{20}{4} , \dfrac{-16}{4} \right) = (5,-4)$